Problem 1776

Solution:

Let $a_{1}, a_{2}, a_{3}$ be the numbers of isosceles triangles whose vertices show exactly $1$, $2$, or $3$ colors, respectively. We assume that $a_{3}=0$ holds. Let the colors be red, green, and blue, where $r, g$, and $b$ denote the (odd) number of vertices colored in each respective color. We now determine in two ways the number $a$ of pairs $(\Delta, v)$, where $\Delta$ is an isosceles triangle with more than one vertex color and $v$ is a side of this triangle whose endpoints are colored with different colors.

Since $a_{3}=0$, the vertices of such a triangle must show exactly two colors, one of which belongs to two vertices that are each endpoints of a side $v$. Thus each triangle contributes two pairs, and it follows that $a=2 a_{2}$.

For any two vertices $A$ and $B$, there are exactly three distinct vertices $C$ that form an isosceles triangle with $A$ and $B$: either $AB=AC$ or $AB=BC$ or $AC=BC$. None of these possibilities can coincide, since otherwise $ABC$ would be equilateral and $n$ would be divisible by $3$. The case $AC=BC$ exists because $n$ is odd, and therefore the perpendicular bisector of $AB$ passes through exactly one further vertex. Hence, starting from two differently colored vertices $A$ and $B$, we have $a=3(rg+gb+br)$. This term is odd by assumption, contradicting $a=2 a_{2}$. Therefore $a_{3} \neq 0$ must hold.